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Homework Help: Electric Field Plastic Rod Problem

  1. Sep 14, 2008 #1

    Can someone help with this? A long, thin, nonconducting plastic rod is bent into a circular loop that has radius a. Between the ends of the rod a short gap of length l, where l << a remains. A positive test charge of magnitude Q is evenly distributed on the loop. What's the magnitude tof E at the center of the loop?

    I don't think we can use Gauss' law for this, because the rod is not connected? So does this mean that we should use dE = k*dq/r^2, and dq = ads, and ds = ad(theta) and integrate? But instead of theta being from 0 to 2*pi, we need to find the theta that is represented by the l, which is I guess l/a for very small theta.

    Also, I read in the book that the electric field is discontinuous at any location with an infinite volume charge density. Can someone explain that? How can there be an infinite volume charge density? Is it like when R goes to zero or something?

  2. jcsd
  3. Sep 14, 2008 #2


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    why don't you want to do the integral?
    sounds like you would know the field if integral ran from 0 to 2pi ;
    You know that 0 to theta is the same as (0 to 2pi) -minus- (2pi to theta)

    Nature doesn't really have INFINITE charge density,
    but it can be pretty dense in a nucleus (16E-18 C / 3fm^3)
  4. Sep 14, 2008 #3


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    For the first one, it seems you aren't given any numbers. So, definitely your integral would look very awkward if you were to do it. So I suggest that you think of it as a complete loop. Then the E-field in the centre is 0. When you remove a small arc portion of the loop, there is nothing to cancel out the E-field due to a small of arc of similar shap and size on the opposite side of the circle. Just find the E-field due to that small portion and you're done.

    As for the second, if you know that the volume charge density is also given by [tex]\nabla \cdot \mathbf{D} = p_v[/tex], you can easily see what this means when the RHS approaches infinity. What does this mean for the LHS?
  5. Sep 15, 2008 #4

    So for the first one, I find the normal equation for a loop of length 2*pi*a + l, and figure out what the new radius would be. Which would be (2*pi*a + l)/2*pi. And then I would use the equation for the loop and then figure out the portion of the field that length l represents?

    And for the second one, do you mean that the delta is the change in density? So does this mean that the distance from the charge to whatever you are measuring goes to 0?

  6. Sep 15, 2008 #5
    Sorry, I think the r is still a, so I just find the portion of the loop with length l and the field that it represents. But the r is still a. I think that is good?
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